The area of the region, enclosed by the circle x2 + y2 = 2 which is not common to the region bounded by the parabola y2 = x and the straight line y = x, is:

Area Under the Curves PYQ — JEE Main 2020 · 7 Jan · Shift 1 · Q42 with Solution

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Correct answer: D

Interpret the regions

We need the area of the part inside the circle $x^2+y^2=2$ but not common to the region bounded by $y^2=x \quad \text{and} \quad y=x.$

So required area = $\text{Area of circle} - \text{Area common to circle and the bounded region.}$

Since the bounded region between $y^2=x$ and $y=x$ lies entirely inside the circle in the first quadrant, the common area is exactly that bounded area.

Find the region bounded by $y^2=x$ and $y=x$

Write the line as: $x=y.$ The parabola is: $x=y^2.$

Their points of intersection satisfy $y=y^2$ so $y(y-1)=0 \implies y=0,1.$ Thus intersections are: $(0,0),\,(1,1).$

For $0\le y\le 1$ , $x_{\text{right}}=y,\qquad x_{\text{left}}=y^2.$ Hence the enclosed area is $\int_0^1 (y-y^2)\,dy.$

Check that this region lies inside the circle

On the line segment from $(0,0)$ to $(1,1)$ , $x^2+y^2=2y^2\le 2.$ On the parabola from $(0,0)$ to $(1,1)$ , $x=y^2 \implies x^2+y^2=y^4+y^2 \le 1+1=2.$ So the whole bounded region lies inside the circle.

Therefore, $\text{required area} = \text{area of circle} - \text{area between curves}.$

Area of the circle

Given $x^2+y^2=2,$ so radius is $r=\sqrt{2}.$ Therefore area of the circle is $\pi r^2 = \pi(2)=2\pi.$

Area between parabola and line

=\left[\frac{y^2}{2}-\frac{y^3}{3}\right]_0^1 =\frac12-\frac13 =\frac16.$$ --- 6. **Required area** $$2\pi-\frac16 = \frac{12\pi-1}{6}.$$ So the required area is $$\boxed{\frac16(12\pi-1)}.$$ --- 7. **Match with options** This is **Option D**. --- 8. **Comparison with stored answer** Stored correct answer: **D** Our derived answer: **D** They agree.

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View written solutionFree

Correct answer: D

Interpret the regions

We need the area of the part inside the circle $x^2+y^2=2$ but not common to the region bounded by $y^2=x \quad \text{and} \quad y=x.$

So required area = $\text{Area of circle} - \text{Area common to circle and the bounded region.}$

Since the bounded region between $y^2=x$ and $y=x$ lies entirely inside the circle in the first quadrant, the common area is exactly that bounded area.

Find the region bounded by $y^2=x$ and $y=x$

Write the line as: $x=y.$ The parabola is: $x=y^2.$

Their points of intersection satisfy $y=y^2$ so $y(y-1)=0 \implies y=0,1.$ Thus intersections are: $(0,0),\,(1,1).$

For $0\le y\le 1$ , $x_{\text{right}}=y,\qquad x_{\text{left}}=y^2.$ Hence the enclosed area is $\int_0^1 (y-y^2)\,dy.$

Check that this region lies inside the circle

Therefore, $\text{required area} = \text{area of circle} - \text{area between curves}.$

Area of the circle

Given $x^2+y^2=2,$ so radius is $r=\sqrt{2}.$ Therefore area of the circle is $\pi r^2 = \pi(2)=2\pi.$

Area between parabola and line

=\left[\frac{y^2}{2}-\frac{y^3}{3}\right]_0^1 =\frac12-\frac13 =\frac16.$$ --- 6. **Required area** $$2\pi-\frac16 = \frac{12\pi-1}{6}.$$ So the required area is $$\boxed{\frac16(12\pi-1)}.$$ --- 7. **Match with options** This is **Option D**. --- 8. **Comparison with stored answer** Stored correct answer: **D** Our derived answer: **D** They agree.

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